1. ECE 3455: Electronics Diode Physics: A Brief Tour

2. Figure 3.7 The i– v characteristic of a silicon junction diode. Figure 3.8 The diode i– v relationship with some scales expanded and others compressed in order to reveal details. Current-Voltage Characteristics We’ll try to justify this… …but we’ll leave this alone until Solid State Devices.

3. Semiconductors of practical interest are generally crystalline, meaning their atoms are arranged in periodic arrays. This figure shows the “diamond” lattice: if the atoms are carbon, this is diamond; if they are Si or Ge we get the corresponding semiconductors (silicon or germanium). If the atoms alternate between Ga and As, we get GaAs (gallium arsenide), one of the important optically sensitive semiconductors (LEDs, photo- detectors).

4. Conduction (i.e., current flow) can take place via movement of either electrons or holes when a voltage (i.e., and electric field) is applied. A hole is the absence of an electron and “moves” when successive electrons take its place. Often, either electrons or holes dominate the current; the other current component is then very small.

5. We can add either electrons or holes to increase the conductivity. This process is called “doping” and involves intentional addition of impurities to the semiconductor. Electrons are added to Si by doping with a Group V (periodic table) element, typically P but also Sb. These have a fifth valence electron which is very loosely bound to the lattice and can move if a voltage is applied. The P or Sb atoms are donors. Holes are added to Si by doping with a Group III element, typically B but also Al. The B or Al atoms are acceptors.

6. n-typep-type Electrons have been added via doping with donors. Holes have been added via doping with acceptors. Here is a pn junction diode… …and the corresponding circuit symbol. n-typep-type Let’s make a diode!! Ingredients: mostly holesIngredients: mostly electrons

7. Why do electrons (or holes, for that matter) move in a semiconductor in the first place? There are two possibilities… Diffusion: if there are more electrons (or holes) in one place than another, they will tend to diffuse to “even out” the concentration. Drift: if an electric field is applied (by applying a voltage), electrons will move toward more positive potential. Due to the concentration gradient, electrons will tend to move toward the right.  - +

8. n-typep-type So when p and n meet… …electrons here diffuse to p-side …holes here diffuse to n-side n-typep-type -+ But that sets up a positively charged region on the n-type side and a negatively charged region on the p-type side... (but see note below) …which generates an electric field that opposes further diffusion.  The charge shown in the lower figure is not due to electrons and holes. Once the electrons get to the p-type side, they disappear when they find holes there (the electrons and holes recombine); the same goes for holes on the n-type side. The charge is due to donors (on the n-side), which become positively charged when they lose their electrons (because they move to the other side!!), and to acceptors (p-side), which become negatively charged when they lose their holes. The donors and acceptors do not move; they are bound to the lattice, and are responsible for the charged regions shown.

9. n-typep-type -+ In fact, electrons and holes never stop moving to the other side via diffusion, and the electric field never stops pulling them back to where they started. But eventually we reach a balance between diffusion and drift; this is equilibrium. (We are assuming there is no applied voltage here.) Electron diffusion Electron drift Hole drift Hole diffusion We’re here 

10. Another way to look at this is the following: at equilibrium, there is an energy barrier keeping more (net) electrons from getting to the p-type side. But we could get another electron to diffuse over if we add energy somehow. e- Electron energy (Joules) If we add energy, this electron can diffuse to the p-side. Similar arguments can be made for holes. e- Instead of adding energy, could we reduce the energy barrier? Now it takes less energy to get the electron over the barrier. The number of electrons that get over the hill increases exponentially with voltage. p n weak electron extra energy diffusion Electron energy (Joules)

11. How can we reduce the barrier? We can apply a voltage!!! e-VaVa ff Note that V a generates an electric field that is opposite in sign to  o. So the net field reduces to  f, and the barrier to diffusion drops. What if we reverse the sign of V a ? The increase in the number of electrons that can get over the barrier is exponential: the current goes as e Va/VT, where VT is the ‘thermal voltage”

12. Now V a generates an electric field that is in the same direction as  o. So the net field increases to  r, and the barrier gets big. There’s no diffusion now. e- VaVa rr But what’s been happening to the drift current?

13. The drift current doesn’t change very much with field, because of the Wile E. Coyote Theorem: regardless of how big the cliff is (no matter what voltage is applied), Wile E. will either fall, or not. Electrons that do fall contribute to the drift current just the same, whether the field is big or small. Wile E. Coyote Theorem small field big field rr ff ACME Red Arrows (Box of 25)

14. Now we put this all together: e- ff rr Reverse bias: Diffusion is “off”: the only current is the drift current, which is small and negative. Forward bias: the diffusion current, which is positive, increases exponentially with voltage. The drift current is negligible. v < 0 v > 0

15. What about that threshold voltage thing?* n-typep-type -+  It turns out that the equilibrium electric field corresponds to a voltage of about 0.8 V or so. When the applied voltage is about 0.6 – 0.7 V, we start to see a significant current**. So we can approximate a threshold voltage Vth as being about 0.7 V in many cases. **In fact, the applied voltage does not all go to changing  . Some of it is dropped across the equivalent resistance of the n- and p-type regions. So at Va = 0.7 V, something less than that is appearing across the barrier and reducing it enough to generate a noticeable current. *Don’t confuse the threshold voltage (Vth or Vf in your textbook) with the thermal voltage VT; these are not the same thing!!